If ( 3, 0 ), ( 2, a) and ( b, 6 ) are the vertices of a ∆ ABC whose centroid is (2, 5), find the values

of a and b.

Q) Show that the points (a, b+c), (b, c +a) and (c, a+b) lie on the straight line

(1)

Given: Vertices of ∆ABC or (3, 0), (2, a) and (b, 6) and vertices of centroid = (2, 5)

⇒5 + b = 3 × 2 = 6  and a + 6 = 5 × 3 = 15

⇒b = 6 – 5 = 1  and a = 15 – 6 = 9

Hence the values of a and b are 9 and 1

(2)

For points (a, b + c), (b, c + a), and (c, a + b) to be on the straight line the area of ∆ formed by the given points will be 0.

⇒ a [(c + a) – (a + b) ] + b [(a + b) – (b + c) ] + c [(b + c) – (c + a) ] = 0

⇒ a (c – b) + b (a – c) + c (b – a) = 0

⇒ ac – ab + ab – bc + bc – ac = 0

⇒ 0 = 0

Hence the given points are collinears